My Math Forum

My Math Forum (http://mymathforum.com/math-forums.php)
-   Calculus (http://mymathforum.com/calculus/)
-   -   Question regarding a critical point formula (http://mymathforum.com/calculus/37630-question-regarding-critical-point-formula.html)

MFP August 20th, 2013 05:26 AM

Question regarding a critical point formula
 
Hey everyone,

I fear that on my final I'll be given an equation that ressembles , and I won't be able to find the critical points.

I know that when you have the form , you can use .

Is there such a method for ?

agentredlum August 20th, 2013 05:58 AM

Re: Question regarding a critical point formula
 
There is a cubic formula

http://en.m.wikipedia.org/wiki/Cubic_function

But it is a bit complicated to use so even if you get a cubic function it is better to use the techniques of precalculus to determine rational roots (if any).

:D

MFP August 20th, 2013 09:47 AM

Re: Question regarding a critical point formula
 
ouch, I don't think I'll be able to memorize that formula in time haha. Alright, I'll use other methods. Thanks agentredlum!

John Marsh August 21st, 2013 05:19 AM

Re: Question regarding a critical point formula
 
"Solution: Given ax+bx+cx+d
let f(x)= ax+bx+cx+d
f'(x)=3ax+2bx+c,where f'(x)=0
then there are two critical point ,and at least two real roots
there for we can use the formula x=[-2bv(4b-12ac)]/6a"

greg1313 August 21st, 2013 10:13 AM

Re: Question regarding a critical point formula
 
Quote:

Originally Posted by John Marsh
then there are two critical point ,and at least two real roots

Not true. Consider x + x + 5x + 3, for example.

HallsofIvy August 21st, 2013 10:48 AM

Re: Question regarding a critical point formula
 
Quote:

Originally Posted by John Marsh
"Solution: Given ax+bx+cx+d
let f(x)= ax+bx+cx+d
f'(x)=3ax+2bx+c,where f'(x)=0
then there are two critical point ,and at least two real roots
there for we can use the formula x=[-2bv(4b-12ac)]/6a"

I don't believe that was the question- although I thought it was at first, myself.

IF the original function were a cubic, then the derivative would be a quadratic but the OP ways he/she solved a quadratic before.
I would interpret this as asking what to do if the derivative itself is a cubic polynomial.

MFP, I doubt that you would be expected to know the cubic formula. IF the derivative turns out to be cubic, try simple integer values for x- bet on it being easily factorable.

slipjack August 21st, 2013 03:42 PM

Let c = p - q. Assume a ? 0, b, p and q are real numbers.
However if we have ax + bx + p = q then we have ax + bx + p - q = 0.
The typical quadratic formula would be .
To prove the quadratic formula is quite easy e.g. divide both sides by a then . . .
The quadratic formula is true if and only if ax + bx + c = 0.


All times are GMT -8. The time now is 11:04 PM.

Copyright © 2017 My Math Forum. All rights reserved.